LeetCode 235. Lowest Common Ancestor of a Binary Search Tree (二叉搜索树最近的共同祖先)

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

 


题目标签:Tree
  这道题目给了我们一个二叉搜索树,和两个点,让我们找到这两个点的最近的共同祖先。什么是最近的共同祖先的呢?来看看题目给的例子,如果给的两个点是3和5, 那么它们的LCA就是4;如果给的是3和9,那么它们的LCA就是6。这道题目我们要根据BST的特性来做,BST的每一个点,它的左边的孩子都比它小,右边的孩子都比它大。
  所以根据上述的这一点,我们每遍历一个点,就来判断一下要继续往左边走,还是往右边走,还是找到了祖先。
  这里有三种情况:(对于每一个遍历点)
    1. 拿p和q之间大的那一个 叫t-max, 和 遍历的点node比较一下,如果t-max 比node val 小的话,说明p和q 都在node 的左边,那么我们要继续走到左边取找;
    2. 拿p和q之间小的那一个,t-min, 和遍历的点node比较一下,如果t-min 比node val 大的话,说明p和q都在node 的右边,那么我们要继续走到右边取找;
    3. 剩下的情况就说明,p和q 在 node 的左右两边,或者是node 是p或者q, 另外一个点在node 的一边。 这种情况就是找到了祖先,把这一个node 返回。
 
 
 

Java Solution:

Runtime beats 62.62% 

完成日期:07/04/2017

关键词:Tree (BST); LCA (lowest common ancestor)

关键点:只遍历左边或者右边根据BST的特性

 

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution 
11 {
12     
13     
14     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) 
15     {
16        if(root == null)
17             return null;
18         
19         // go into left child if the greater from p and q is smaller than root;
20         if(Math.max(p.val, q.val) < root.val)
21             return lowestCommonAncestor(root.left, p, q);
22         
23         // go into right child if the smaller from p and q is greater than root;
24         if(Math.min(p.val, q.val) > root.val)
25             return lowestCommonAncestor(root.right, p, q);
26         
27         // if p and q are not at the left side, and also not at the right side
28         // meaning p and q are at the both side or root is p or q , another one is at one side.
29         return root;
30     }
31     
32     
33 }

参考资料:

http://www.cnblogs.com/grandyang/p/4640572.html

 

LeetCode 算法题目列表 - LeetCode Algorithms Questions List

posted @ 2017-07-05 01:02  Jimmy_Cheng  阅读(268)  评论(0编辑  收藏  举报